Download Introduction to Combinatorics on Words

Combinatorics on Words:
An Introduction
Jeffrey Shallit
School of Computer Science
University of Waterloo
Waterloo, Ontario N2L 3G1
Canada
[email protected]
http://www.cs.uwaterloo.ca/~shallit
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The Main Themes
periodicity
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The Main Themes
periodicity
patterns and pattern avoidance
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The Main Themes
periodicity
patterns and pattern avoidance
equations in words
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The Main Themes
periodicity
patterns and pattern avoidance
equations in words
infinite words and their properties
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Some notation
Σ - a finite nonempty set of symbols - the alphabet
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Some notation
Σ - a finite nonempty set of symbols - the alphabet
word - a finite or infinite list of symbols chosen from Σ
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Some notation
Σ - a finite nonempty set of symbols - the alphabet
word - a finite or infinite list of symbols chosen from Σ
Σ∗ - set of all finite words
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Some notation
Σ - a finite nonempty set of symbols - the alphabet
word - a finite or infinite list of symbols chosen from Σ
Σ∗ - set of all finite words
Σ+ - set of all finite nonempty words
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Some notation
Σ - a finite nonempty set of symbols - the alphabet
word - a finite or infinite list of symbols chosen from Σ
Σ∗ - set of all finite words
Σ+ - set of all finite nonempty words
Σω - set of all (right-)infinite words
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Some notation
Σ - a finite nonempty set of symbols - the alphabet
word - a finite or infinite list of symbols chosen from Σ
Σ∗ - set of all finite words
Σ+ - set of all finite nonempty words
Σω - set of all (right-)infinite words
Σ∞ = Σ∗ ∪ Σω
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More notation
the empty word: ²
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More notation
the empty word: ²
w = a 1 a2 · · · a n
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More notation
the empty word: ²
w = a 1 a2 · · · a n
w [i] := ai , w [i..j] := ai ai+1 · · · aj
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More notation
the empty word: ²
w = a 1 a2 · · · a n
w [i] := ai , w [i..j] := ai ai+1 · · · aj
w = a 0 a1 a2 · · ·
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More notation
the empty word: ²
w = a 1 a2 · · · a n
w [i] := ai , w [i..j] := ai ai+1 · · · aj
w = a 0 a1 a2 · · ·
x ω = xxx · · ·
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More notation
the empty word: ²
w = a 1 a2 · · · a n
w [i] := ai , w [i..j] := ai ai+1 · · · aj
w = a 0 a1 a2 · · ·
x ω = xxx · · ·
ultimately periodic: z = xy ω
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More notation
the empty word: ²
w = a 1 a2 · · · a n
w [i] := ai , w [i..j] := ai ai+1 · · · aj
w = a 0 a1 a2 · · ·
x ω = xxx · · ·
ultimately periodic: z = xy ω
n
Operations: concatenation, raising to powers
reversal x R
xn
z }| {
= xx · · · x, x 0 = ²,
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More notation
the empty word: ²
w = a 1 a2 · · · a n
w [i] := ai , w [i..j] := ai ai+1 · · · aj
w = a 0 a1 a2 · · ·
x ω = xxx · · ·
ultimately periodic: z = xy ω
n
Operations: concatenation, raising to powers
reversal x R
xn
z }| {
= xx · · · x, x 0 = ²,
prefix, suffix, factor, subword
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Algebraic framework
semigroup: concatenation is multiplication, associative
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Algebraic framework
semigroup: concatenation is multiplication, associative
monoid: semigroup + identity element (²)
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Algebraic framework
semigroup: concatenation is multiplication, associative
monoid: semigroup + identity element (²)
free monoid: no relations among elements
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Algebraic framework
semigroup: concatenation is multiplication, associative
monoid: semigroup + identity element (²)
free monoid: no relations among elements
group: add inverses of elements a−1
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Periodicity
words - fundamentally noncommutative
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Periodicity
words - fundamentally noncommutative
casebook 6= bookcase
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Periodicity
words - fundamentally noncommutative
casebook 6= bookcase
When do words commute?
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Periodicity
words - fundamentally noncommutative
casebook 6= bookcase
When do words commute?
Here are two words that “almost” commute:
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Periodicity
words - fundamentally noncommutative
casebook 6= bookcase
When do words commute?
Here are two words that “almost” commute:
w = 01010 and x = 01011010
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Periodicity
words - fundamentally noncommutative
casebook 6= bookcase
When do words commute?
Here are two words that “almost” commute:
w = 01010 and x = 01011010
wx
= 0101001011010
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Periodicity
words - fundamentally noncommutative
casebook 6= bookcase
When do words commute?
Here are two words that “almost” commute:
w = 01010 and x = 01011010
wx
= 0101001011010
xw
= 0101101001010
By the way, this raises the question: can the Hamming distance
between wx and xw be 1? It can’t; there is a one-line proof.
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What are the solutions to x 2 = y 3 in words?
- Over N, x 2 = y 3 iff x is a cube and y is a square
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What are the solutions to x 2 = y 3 in words?
- Over N, x 2 = y 3 iff x is a cube and y is a square
- Suggests what solutions of x i = y j will look like over words
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What are the solutions to x 2 = y 3 in words?
- Over N, x 2 = y 3 iff x is a cube and y is a square
- Suggests what solutions of x i = y j will look like over words
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First Theorem of Lyndon-Schützenberger
Theorem
Let x, y ∈ Σ+ . Then the following three conditions are equivalent:
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First Theorem of Lyndon-Schützenberger
Theorem
Let x, y ∈ Σ+ . Then the following three conditions are equivalent:
(1) xy = yx;
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First Theorem of Lyndon-Schützenberger
Theorem
Let x, y ∈ Σ+ . Then the following three conditions are equivalent:
(1) xy = yx;
(2) There exist z ∈ Σ+ and integers k, l > 0 such that x = z k and
y = zl;
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First Theorem of Lyndon-Schützenberger
Theorem
Let x, y ∈ Σ+ . Then the following three conditions are equivalent:
(1) xy = yx;
(2) There exist z ∈ Σ+ and integers k, l > 0 such that x = z k and
y = zl;
(3) There exist integers i, j > 0 such that x i = y j .
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Second Theorem of Lyndon-Schützenberger
Under what conditions can a string have a nontrivial proper prefix
and suffix that are identical?
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Second Theorem of Lyndon-Schützenberger
Under what conditions can a string have a nontrivial proper prefix
and suffix that are identical?
Examples in English: reader — begins and ends with r
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Second Theorem of Lyndon-Schützenberger
Under what conditions can a string have a nontrivial proper prefix
and suffix that are identical?
Examples in English: reader — begins and ends with r
alfalfa — which begins and ends with alfa
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Second Theorem of Lyndon-Schützenberger
Under what conditions can a string have a nontrivial proper prefix
and suffix that are identical?
Examples in English: reader — begins and ends with r
alfalfa — which begins and ends with alfa
The answer is given by the following theorem.
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Second Theorem of Lyndon-Schützenberger
Under what conditions can a string have a nontrivial proper prefix
and suffix that are identical?
Examples in English: reader — begins and ends with r
alfalfa — which begins and ends with alfa
The answer is given by the following theorem.
Theorem
Let x, y , z ∈ Σ+ . Then xy = yz if and only if there exist u ∈ Σ+ ,
v ∈ Σ∗ , and an integer e ≥ 0 such that x = uv , z = vu, and
y = (uv )e u = u(vu)e .
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Primitive words
We say a word x is a power if it can be expressed as x = y n for
some y 6= ², n ≥ 2.
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Primitive words
We say a word x is a power if it can be expressed as x = y n for
some y 6= ², n ≥ 2.
A nonpower is called primitive.
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Primitive words
We say a word x is a power if it can be expressed as x = y n for
some y 6= ², n ≥ 2.
A nonpower is called primitive.
Every nonempty word can be written uniquely in the form x k
where x is primitive and k ≥ 1.
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Primitive words
We say a word x is a power if it can be expressed as x = y n for
some y 6= ², n ≥ 2.
A nonpower is called primitive.
Every nonempty word can be written uniquely in the form x k
where x is primitive and k ≥ 1.
Enumeration: there are exactly
X
µ(d)k n/d
d |n
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Primitive words
We say a word x is a power if it can be expressed as x = y n for
some y 6= ², n ≥ 2.
A nonpower is called primitive.
Every nonempty word can be written uniquely in the form x k
where x is primitive and k ≥ 1.
Enumeration: there are exactly
X
µ(d)k n/d
d |n
primitive words of length n over a k-letter alphabet. Here µ is the
Möbius function and the sum is over the divisors of n.
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Primitive words
We say a word x is a power if it can be expressed as x = y n for
some y 6= ², n ≥ 2.
A nonpower is called primitive.
Every nonempty word can be written uniquely in the form x k
where x is primitive and k ≥ 1.
Enumeration: there are exactly
X
µ(d)k n/d
d |n
primitive words of length n over a k-letter alphabet. Here µ is the
Möbius function and the sum is over the divisors of n.
Open question: is the set of primitive binary words a CFL?
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Conjugates
A word w is a conjugate of a word x if w can be obtained from x
by cyclically shifting the letters.
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Conjugates
A word w is a conjugate of a word x if w can be obtained from x
by cyclically shifting the letters.
For example, the English word enlist is a conjugate of listen.
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Conjugates
A word w is a conjugate of a word x if w can be obtained from x
by cyclically shifting the letters.
For example, the English word enlist is a conjugate of listen.
A conjugate of a k-th power is a k-th power of a conjugate.
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Conjugates
A word w is a conjugate of a word x if w can be obtained from x
by cyclically shifting the letters.
For example, the English word enlist is a conjugate of listen.
A conjugate of a k-th power is a k-th power of a conjugate.
Every primitive word has an unbordered conjugate.
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Conjugates
A word w is a conjugate of a word x if w can be obtained from x
by cyclically shifting the letters.
For example, the English word enlist is a conjugate of listen.
A conjugate of a k-th power is a k-th power of a conjugate.
Every primitive word has an unbordered conjugate.
Lyndon word: lexicographically least among all its conjugates
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Conjugates
A word w is a conjugate of a word x if w can be obtained from x
by cyclically shifting the letters.
For example, the English word enlist is a conjugate of listen.
A conjugate of a k-th power is a k-th power of a conjugate.
Every primitive word has an unbordered conjugate.
Lyndon word: lexicographically least among all its conjugates
Theorem: Every finite word has a unique factorization as the
product of Lyndon words w1 w2 · · · wn , where w1 ≥ w2 ≥ w3 · · · wn .
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Fractional powers
We say a word w is a p/q power, for integers p ≥ q ≥ 1, if
w = x bp/qc x 0
for a prefix x 0 of x such that |w |/|x| = p/q.
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Fractional powers
We say a word w is a p/q power, for integers p ≥ q ≥ 1, if
w = x bp/qc x 0
for a prefix x 0 of x such that |w |/|x| = p/q.
For example, the French word entente is a 7/3-power, as it can
be written as (ent)2 e.
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Fractional powers
We say a word w is a p/q power, for integers p ≥ q ≥ 1, if
w = x bp/qc x 0
for a prefix x 0 of x such that |w |/|x| = p/q.
For example, the French word entente is a 7/3-power, as it can
be written as (ent)2 e.
The German word schematische is a 3/2 power.
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Fractional powers
We say a word w is a p/q power, for integers p ≥ q ≥ 1, if
w = x bp/qc x 0
for a prefix x 0 of x such that |w |/|x| = p/q.
For example, the French word entente is a 7/3-power, as it can
be written as (ent)2 e.
The German word schematische is a 3/2 power.
If w = x bp/qc x 0 is a p/q power, then we call x a period of w .
Often the word period is used to refer to |x|.
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Fractional powers
We say a word w is a p/q power, for integers p ≥ q ≥ 1, if
w = x bp/qc x 0
for a prefix x 0 of x such that |w |/|x| = p/q.
For example, the French word entente is a 7/3-power, as it can
be written as (ent)2 e.
The German word schematische is a 3/2 power.
If w = x bp/qc x 0 is a p/q power, then we call x a period of w .
Often the word period is used to refer to |x|.
If a word w is a p/q > 1 power, then it begins and ends with some
nonempty string. Such a string is also called bordered. Otherwise
it is unbordered.
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Unbordered words
The unbordered words play the same role for fractional powers as
the primitive words do for ordinary powers.
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Unbordered words
The unbordered words play the same role for fractional powers as
the primitive words do for ordinary powers.
Enumeration of unbordered words is more challenging and there is
no simple closed form.
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Unbordered words
The unbordered words play the same role for fractional powers as
the primitive words do for ordinary powers.
Enumeration of unbordered words is more challenging and there is
no simple closed form.
However, there are asymptotically ck k n such words, where ck is a
constant that tends to 1 as k tends to ∞.
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Duval’s conjecture
Let µ(w ) be the length of the longest unbordered factor of w .
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Duval’s conjecture
Let µ(w ) be the length of the longest unbordered factor of w .
Let p(w ) be the length of the longest period of w .
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Duval’s conjecture
Let µ(w ) be the length of the longest unbordered factor of w .
Let p(w ) be the length of the longest period of w .
Duval conjectured that if |w | ≥ 3µ(w ), then µ(w ) = p(w ).
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Duval’s conjecture
Let µ(w ) be the length of the longest unbordered factor of w .
Let p(w ) be the length of the longest period of w .
Duval conjectured that if |w | ≥ 3µ(w ), then µ(w ) = p(w ).
This was proved by Harju & Nowotka, and S. Holub. The result
has been improved to |w | ≥ 3µ(w ) − 2 =⇒ µ(w ) = p(w ).
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The Fine-Wilf theorem
Theorem
Let w and x be nonempty words. Let y ∈ w {w , x}ω and
z ∈ x{w , x}ω . Then the following conditions are equivalent:
(a) y and z agree on a prefix of length at least
|w | + |x| − gcd(|w |, |x|);
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The Fine-Wilf theorem
Theorem
Let w and x be nonempty words. Let y ∈ w {w , x}ω and
z ∈ x{w , x}ω . Then the following conditions are equivalent:
(a) y and z agree on a prefix of length at least
|w | + |x| − gcd(|w |, |x|);
(b) wx = xw ;
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The Fine-Wilf theorem
Theorem
Let w and x be nonempty words. Let y ∈ w {w , x}ω and
z ∈ x{w , x}ω . Then the following conditions are equivalent:
(a) y and z agree on a prefix of length at least
|w | + |x| − gcd(|w |, |x|);
(b) wx = xw ;
(c) y = z.
(c) =⇒ (a): Trivial.
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The Fine-Wilf theorem
Theorem
Let w and x be nonempty words. Let y ∈ w {w , x}ω and
z ∈ x{w , x}ω . Then the following conditions are equivalent:
(a) y and z agree on a prefix of length at least
|w | + |x| − gcd(|w |, |x|);
(b) wx = xw ;
(c) y = z.
(c) =⇒ (a): Trivial.
We’ll prove (a) =⇒ (b) and (b) =⇒ (c).
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Fine-Wilf: The proof
Proof.
(a) y and z agree on a prefix of length at least
|w | + |x| − gcd(|w |, |x|) =⇒ (b) wx = xw
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Fine-Wilf: The proof
Proof.
(a) y and z agree on a prefix of length at least
|w | + |x| − gcd(|w |, |x|) =⇒ (b) wx = xw
We prove the contrapositive. Suppose wx 6= xw .
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Fine-Wilf: The proof
Proof.
(a) y and z agree on a prefix of length at least
|w | + |x| − gcd(|w |, |x|) =⇒ (b) wx = xw
We prove the contrapositive. Suppose wx 6= xw .
Then we prove that y and z differ at a position
≤ |w | + |x| − gcd(|w |, |x|).
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Fine-Wilf: The proof
Proof.
(a) y and z agree on a prefix of length at least
|w | + |x| − gcd(|w |, |x|) =⇒ (b) wx = xw
We prove the contrapositive. Suppose wx 6= xw .
Then we prove that y and z differ at a position
≤ |w | + |x| − gcd(|w |, |x|).
The proof is by induction on |w | + |x|.
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Fine-Wilf: The proof
Proof.
(a) y and z agree on a prefix of length at least
|w | + |x| − gcd(|w |, |x|) =⇒ (b) wx = xw
We prove the contrapositive. Suppose wx 6= xw .
Then we prove that y and z differ at a position
≤ |w | + |x| − gcd(|w |, |x|).
The proof is by induction on |w | + |x|.
The base case is |w | + |x| = 2. Then |w | = |x| = 1, and
|w | + |x| − gcd(|w |, |x|) = 1. Since wx 6= xw , we must have
w = a, x = b with a 6= b. Then y and z differ at the first position.
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Now assume the result is true for |w | + |x| < k.
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Now assume the result is true for |w | + |x| < k.
We prove it for |w | + |x| = k.
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Now assume the result is true for |w | + |x| < k.
We prove it for |w | + |x| = k.
If |w | = |x| then y and z must disagree at the |w |’th position or
earlier, for otherwise w = x and wx = xw ; since
|w | ≤ |w | + |x| − gcd(|w |, |x|) = |w |, the result follows.
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Now assume the result is true for |w | + |x| < k.
We prove it for |w | + |x| = k.
If |w | = |x| then y and z must disagree at the |w |’th position or
earlier, for otherwise w = x and wx = xw ; since
|w | ≤ |w | + |x| − gcd(|w |, |x|) = |w |, the result follows.
So, without loss of generality, assume |w | < |x|.
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Now assume the result is true for |w | + |x| < k.
We prove it for |w | + |x| = k.
If |w | = |x| then y and z must disagree at the |w |’th position or
earlier, for otherwise w = x and wx = xw ; since
|w | ≤ |w | + |x| − gcd(|w |, |x|) = |w |, the result follows.
So, without loss of generality, assume |w | < |x|.
If w is not a prefix of x, then y and z disagree on the |w |’th
position or earlier, and again |w | ≤ |w | + |x| − gcd(|w |, |x|).
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Now assume the result is true for |w | + |x| < k.
We prove it for |w | + |x| = k.
If |w | = |x| then y and z must disagree at the |w |’th position or
earlier, for otherwise w = x and wx = xw ; since
|w | ≤ |w | + |x| − gcd(|w |, |x|) = |w |, the result follows.
So, without loss of generality, assume |w | < |x|.
If w is not a prefix of x, then y and z disagree on the |w |’th
position or earlier, and again |w | ≤ |w | + |x| − gcd(|w |, |x|).
So w is a proper prefix of x.
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Now assume the result is true for |w | + |x| < k.
We prove it for |w | + |x| = k.
If |w | = |x| then y and z must disagree at the |w |’th position or
earlier, for otherwise w = x and wx = xw ; since
|w | ≤ |w | + |x| − gcd(|w |, |x|) = |w |, the result follows.
So, without loss of generality, assume |w | < |x|.
If w is not a prefix of x, then y and z disagree on the |w |’th
position or earlier, and again |w | ≤ |w | + |x| − gcd(|w |, |x|).
So w is a proper prefix of x.
Write x = wt for some nonempty word t.
Now any common divisor of |w | and |x| must also divide
|x| − |w | = |t|, and similarly any common divisor of both |w | and
|t| must also divide |w | + |t| = |x|. So gcd(|w |, |x|) = gcd(|w |, |t|).
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Now wt 6= tw , for otherwise we have wx = wwt = wtw = xw , a
contradiction.
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Now wt 6= tw , for otherwise we have wx = wwt = wtw = xw , a
contradiction.
Then y = ww · · · and z = wt · · · . By induction (since
|w | + |t| < k) w −1 y and w −1 z disagree at position
|w | + |t| − gcd(|w |, |t|) or earlier.
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Now wt 6= tw , for otherwise we have wx = wwt = wtw = xw , a
contradiction.
Then y = ww · · · and z = wt · · · . By induction (since
|w | + |t| < k) w −1 y and w −1 z disagree at position
|w | + |t| − gcd(|w |, |t|) or earlier.
Hence y and z disagree at position
2|w | + |t| − gcd(|w |, |t|) = |w | + |x| − gcd(|w |, |x|) or earlier.
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(b) =⇒ (c): If wx = xw , then by the theorem of
Lyndon-Schützenberger, both w and x are in u + for some word u.
Hence y = u ω = z.
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Finite Sturmian words
The proof also implies a way to get words that optimally “almost
commute”, in the sense that xw and wx should agree on as long a
segment as possible.
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Finite Sturmian words
The proof also implies a way to get words that optimally “almost
commute”, in the sense that xw and wx should agree on as long a
segment as possible.
Theorem
For each m, n ≥ 1 there exist words x, w of length m, n,
respectively, such that xw and wx agree on a prefix of length
m + n − gcd(m, n) − 1 but differ at position m + n − gcd(m, n).
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Finite Sturmian words
The proof also implies a way to get words that optimally “almost
commute”, in the sense that xw and wx should agree on as long a
segment as possible.
Theorem
For each m, n ≥ 1 there exist words x, w of length m, n,
respectively, such that xw and wx agree on a prefix of length
m + n − gcd(m, n) − 1 but differ at position m + n − gcd(m, n).
These are the finite Sturmian words.
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Finite Sturmian words
The proof also implies a way to get words that optimally “almost
commute”, in the sense that xw and wx should agree on as long a
segment as possible.
Theorem
For each m, n ≥ 1 there exist words x, w of length m, n,
respectively, such that xw and wx agree on a prefix of length
m + n − gcd(m, n) − 1 but differ at position m + n − gcd(m, n).
These are the finite Sturmian words.
Many authors have worked on generalizations to multiple periods:
Castelli, Mignosi, & Restivo, Simpson & Tijdeman, Constantinescu
& Ilie, ...
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Patterns and pattern avoidance
The story begins with Axel Thue in 1906.
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Patterns and pattern avoidance
The story begins with Axel Thue in 1906.
He noticed that over a 2-letter alphabet, every word of length ≥ 4
contains a square: either 02 , 12 , (01)2 or (10)2 .
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Patterns and pattern avoidance
The story begins with Axel Thue in 1906.
He noticed that over a 2-letter alphabet, every word of length ≥ 4
contains a square: either 02 , 12 , (01)2 or (10)2 .
But over a 3-letter alphabet, it is possible to create arbitrarily long
words (or — what is equivalent — an infinite word) with no square
factors at all. Such a word is called squarefree.
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Patterns and pattern avoidance
The story begins with Axel Thue in 1906.
He noticed that over a 2-letter alphabet, every word of length ≥ 4
contains a square: either 02 , 12 , (01)2 or (10)2 .
But over a 3-letter alphabet, it is possible to create arbitrarily long
words (or — what is equivalent — an infinite word) with no square
factors at all. Such a word is called squarefree.
The easiest way to construct such a sequence was found by Thue
in 1912 (and rediscovered many times).
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Patterns and pattern avoidance
The story begins with Axel Thue in 1906.
He noticed that over a 2-letter alphabet, every word of length ≥ 4
contains a square: either 02 , 12 , (01)2 or (10)2 .
But over a 3-letter alphabet, it is possible to create arbitrarily long
words (or — what is equivalent — an infinite word) with no square
factors at all. Such a word is called squarefree.
The easiest way to construct such a sequence was found by Thue
in 1912 (and rediscovered many times).
It is based on the Thue-Morse sequence.
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The Thue-Morse morphism
Morphism: a map h from Σ∗ to ∆∗ such that
h(xy ) = h(x)h(y ).
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The Thue-Morse morphism
Morphism: a map h from Σ∗ to ∆∗ such that
h(xy ) = h(x)h(y ).
Thue-Morse morphism: µ(0) = 01; µ(1) = 10.
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The Thue-Morse morphism
Morphism: a map h from Σ∗ to ∆∗ such that
h(xy ) = h(x)h(y ).
Thue-Morse morphism: µ(0) = 01; µ(1) = 10.
If Σ = ∆ then we can iterate h.
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The Thue-Morse morphism
Morphism: a map h from Σ∗ to ∆∗ such that
h(xy ) = h(x)h(y ).
Thue-Morse morphism: µ(0) = 01; µ(1) = 10.
If Σ = ∆ then we can iterate h.
i
We write
hi
z
}|
{
= h(h(h(· · · ))).
99 / 201
Morphic words
If a nonerasing morphism has the property that h(a) = ax, then
iterating h produces an infinite word
hω (a) = a x h(x) h2 (x) h3 (x) · · · .
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Morphic words
If a nonerasing morphism has the property that h(a) = ax, then
iterating h produces an infinite word
hω (a) = a x h(x) h2 (x) h3 (x) · · · .
If we do this with µ we get the Thue-Morse word:
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Morphic words
If a nonerasing morphism has the property that h(a) = ax, then
iterating h produces an infinite word
hω (a) = a x h(x) h2 (x) h3 (x) · · · .
If we do this with µ we get the Thue-Morse word:
t = µω (0) = 0110100110010110 · · · .
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Morphic words
If a nonerasing morphism has the property that h(a) = ax, then
iterating h produces an infinite word
hω (a) = a x h(x) h2 (x) h3 (x) · · · .
If we do this with µ we get the Thue-Morse word:
t = µω (0) = 0110100110010110 · · · .
Also rediscovered by Marston Morse, Max Euwe, Solomon Arshon,
and the Danish composer Per Nørgård.
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Properties of the Thue-Morse word
An overlap is a word of the form axaxa, where a is a single letter
and x is a word.
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Properties of the Thue-Morse word
An overlap is a word of the form axaxa, where a is a single letter
and x is a word.
An example in English is alfalfa (take x = lf).
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Properties of the Thue-Morse word
An overlap is a word of the form axaxa, where a is a single letter
and x is a word.
An example in English is alfalfa (take x = lf).
Thus an overlap is just slightly more than a square.
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Properties of the Thue-Morse word
An overlap is a word of the form axaxa, where a is a single letter
and x is a word.
An example in English is alfalfa (take x = lf).
Thus an overlap is just slightly more than a square.
It is also called a 2+ power.
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Properties of the Thue-Morse word
An overlap is a word of the form axaxa, where a is a single letter
and x is a word.
An example in English is alfalfa (take x = lf).
Thus an overlap is just slightly more than a square.
It is also called a 2+ power.
Theorem
The Thue-Morse word t is overlap-free.
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From overlap-free to squarefree
We can construct a squarefree word from t, as follows:
109 / 201
From overlap-free to squarefree
We can construct a squarefree word from t, as follows:
Count the number of 1’s in t between consecutive 0’s:
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From overlap-free to squarefree
We can construct a squarefree word from t, as follows:
Count the number of 1’s in t between consecutive 0’s:
We get:
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From overlap-free to squarefree
We can construct a squarefree word from t, as follows:
Count the number of 1’s in t between consecutive 0’s:
We get:
2
1
0
2
0
1
2
1
0
1
2
z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{
0 1 0 11 · · ·
0 1 0 11 0 1 0
0 11 0
0 11 0 1 0
112 / 201
From overlap-free to squarefree
We can construct a squarefree word from t, as follows:
Count the number of 1’s in t between consecutive 0’s:
We get:
2
1
0
2
0
1
2
1
0
1
2
z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{ z}|{
0 1 0 11 · · ·
0 1 0 11 0 1 0
0 11 0
0 11 0 1 0
This is squarefree, as a square in this word implies and overlap in
the Thue-Morse word.
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Enumeration of power-free words
How many squarefree words are there?
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Enumeration of power-free words
How many squarefree words are there?
Infinite - countable or uncountable
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Enumeration of power-free words
How many squarefree words are there?
Infinite - countable or uncountable
Finite - polynomially-many or exponentially-many of length n?
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Enumeration of power-free words
How many squarefree words are there?
Infinite - countable or uncountable
Finite - polynomially-many or exponentially-many of length n?
Same question can be asked for the overlap-free words.
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Enumeration of power-free words
How many squarefree words are there?
Infinite - countable or uncountable
Finite - polynomially-many or exponentially-many of length n?
Same question can be asked for the overlap-free words.
For overlap-free words over {0, 1} there is a factorization theorem
of Restivo and Salemi that implies only polynomially-many of
length n.
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Enumeration of power-free words
How many squarefree words are there?
Infinite - countable or uncountable
Finite - polynomially-many or exponentially-many of length n?
Same question can be asked for the overlap-free words.
For overlap-free words over {0, 1} there is a factorization theorem
of Restivo and Salemi that implies only polynomially-many of
length n.
For squarefree words over {0, 1, 2} there are exponentially many.
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Dejean’s Conjecture
Given an alphabet Σ of cardinality k, we can try to find the optimal
(fractional) exponent αk avoidable by infinite words over Σ.
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Dejean’s Conjecture
Given an alphabet Σ of cardinality k, we can try to find the optimal
(fractional) exponent αk avoidable by infinite words over Σ.
For k = 2 we have already seen that overlaps are avoidable and
squares are not. So α2 = 2.
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Dejean’s Conjecture
Given an alphabet Σ of cardinality k, we can try to find the optimal
(fractional) exponent αk avoidable by infinite words over Σ.
For k = 2 we have already seen that overlaps are avoidable and
squares are not. So α2 = 2.
Dejean (1972) showed that α3 = 7/4 and conjectured that
α4 = 7/5 and αk = k/(k − 1) for k ≥ 5.
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Dejean’s Conjecture
Given an alphabet Σ of cardinality k, we can try to find the optimal
(fractional) exponent αk avoidable by infinite words over Σ.
For k = 2 we have already seen that overlaps are avoidable and
squares are not. So α2 = 2.
Dejean (1972) showed that α3 = 7/4 and conjectured that
α4 = 7/5 and αk = k/(k − 1) for k ≥ 5.
This conjecture has been proven by the combined efforts of
Pansiot, Moulin-Ollagnier, Currie & Mohammad-Noori, Carpi,
Currie & Rampersad, and Rao.
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Dejean’s Conjecture
Given an alphabet Σ of cardinality k, we can try to find the optimal
(fractional) exponent αk avoidable by infinite words over Σ.
For k = 2 we have already seen that overlaps are avoidable and
squares are not. So α2 = 2.
Dejean (1972) showed that α3 = 7/4 and conjectured that
α4 = 7/5 and αk = k/(k − 1) for k ≥ 5.
This conjecture has been proven by the combined efforts of
Pansiot, Moulin-Ollagnier, Currie & Mohammad-Noori, Carpi,
Currie & Rampersad, and Rao.
Still open: many other variants of Dejean where the length of the
period is also taken into account.
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More general patterns
Instead of avoiding xx or axaxa, we can try to avoid more general
patterns.
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More general patterns
Instead of avoiding xx or axaxa, we can try to avoid more general
patterns.
“Avoiding the pattern α” means constructing an infinite word x
such that, for all non-erasing morphisms h, the word h(α) is not a
factor of x.
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More general patterns
Instead of avoiding xx or axaxa, we can try to avoid more general
patterns.
“Avoiding the pattern α” means constructing an infinite word x
such that, for all non-erasing morphisms h, the word h(α) is not a
factor of x.
Not all patterns are avoidable — even if the alphabet is arbitrarily
large.
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More general patterns
Instead of avoiding xx or axaxa, we can try to avoid more general
patterns.
“Avoiding the pattern α” means constructing an infinite word x
such that, for all non-erasing morphisms h, the word h(α) is not a
factor of x.
Not all patterns are avoidable — even if the alphabet is arbitrarily
large.
For example - it is impossible to avoid xyx, since every sufficiently
long string z will contain three occurrences of some letter a, say
z = rasatau, and then we can let x = a, y = sat, both x and y are
nonempty.
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Avoiding general patterns
Given a pattern, it is decidable (via Zimin’s algorithm) if it is
avoidable over some alphabet.
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Avoiding general patterns
Given a pattern, it is decidable (via Zimin’s algorithm) if it is
avoidable over some alphabet.
However, we do not have a general procedure to decide if a given
pattern is avoidable over a fixed alphabet.
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Abelian powers
Instead of avoiding xx, we can consider trying to avoid other kinds
of patterns: the so-called abelian powers.
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Abelian powers
Instead of avoiding xx, we can consider trying to avoid other kinds
of patterns: the so-called abelian powers.
An abelian square is a nonempty word of the form xx 0 , where x 0 is
a permutation of x.
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Abelian powers
Instead of avoiding xx, we can consider trying to avoid other kinds
of patterns: the so-called abelian powers.
An abelian square is a nonempty word of the form xx 0 , where x 0 is
a permutation of x.
For example, interessierten is an abelian square in German, as
sierten is a permutation of interes.
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Abelian powers
Instead of avoiding xx, we can consider trying to avoid other kinds
of patterns: the so-called abelian powers.
An abelian square is a nonempty word of the form xx 0 , where x 0 is
a permutation of x.
For example, interessierten is an abelian square in German, as
sierten is a permutation of interes.
In a similar way, we can define abelian cubes as words of the form
xx 0 x 00 where both x 0 and x 00 are permutations of x.
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Abelian powers: summary of results
it is possible to avoid abelian squares over a 4-letter alphabet, and
this is optimal;
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Abelian powers: summary of results
it is possible to avoid abelian squares over a 4-letter alphabet, and
this is optimal;
it is possible to avoid abelian cubes over a 3-letter alphabet, and
this is optimal;
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Abelian powers: summary of results
it is possible to avoid abelian squares over a 4-letter alphabet, and
this is optimal;
it is possible to avoid abelian cubes over a 3-letter alphabet, and
this is optimal;
it is possible to avoid abelian fourth powers over a 2-letter
alphabet, and this is optimal.
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Abelian powers: summary of results
it is possible to avoid abelian squares over a 4-letter alphabet, and
this is optimal;
it is possible to avoid abelian cubes over a 3-letter alphabet, and
this is optimal;
it is possible to avoid abelian fourth powers over a 2-letter
alphabet, and this is optimal.
An open problem: is it possible to avoid, overP
a finiteP
subset of N,
patterns of the form xx 0 where |x| = |x 0 | and
x=
x 0?
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Abelian powers: summary of results
it is possible to avoid abelian squares over a 4-letter alphabet, and
this is optimal;
it is possible to avoid abelian cubes over a 3-letter alphabet, and
this is optimal;
it is possible to avoid abelian fourth powers over a 2-letter
alphabet, and this is optimal.
An open problem: is it possible to avoid, overP
a finiteP
subset of N,
patterns of the form xx 0 where |x| = |x 0 | and
x=
x 0?
Also still open: fractional version of abelian powers
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A generalization of abelian powers
involution: h2 (x) = x for all words x
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A generalization of abelian powers
involution: h2 (x) = x for all words x
involutions can be morphic (h(xy ) = h(x)h(y )) or antimorphic
(h(xy ) = h(y )h(x))
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A generalization of abelian powers
involution: h2 (x) = x for all words x
involutions can be morphic (h(xy ) = h(x)h(y )) or antimorphic
(h(xy ) = h(y )h(x))
One antimorphism of biological interest: reverse string and apply
map A ↔ T , G ↔ C .
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A generalization of abelian powers
involution: h2 (x) = x for all words x
involutions can be morphic (h(xy ) = h(x)h(y )) or antimorphic
(h(xy ) = h(y )h(x))
One antimorphism of biological interest: reverse string and apply
map A ↔ T , G ↔ C .
Can avoid some patterns involving involution, but not others
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Equations in words
Example 1:
abX = Xba.
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Equations in words
Example 1:
abX = Xba.
The only solutions are X ∈ (ab)∗ a.
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Equations in words
Example 1:
abX = Xba.
The only solutions are X ∈ (ab)∗ a.
Example 2:
XaXbY = aXYbX
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Equations in words
Example 1:
abX = Xba.
The only solutions are X ∈ (ab)∗ a.
Example 2:
XaXbY = aXYbX
The solutions are X = ai , Y = (ai b)j ai for i ≥ 0, j ≥ 0.
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Equations in words
Example 3: Fermat’s equation for words: x i y j = z k
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Equations in words
Example 3: Fermat’s equation for words: x i y j = z k
The only solutions for i, j, k ≥ 2 are when x, y , z are all powers of
a third word.
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Equations in words
Example 3: Fermat’s equation for words: x i y j = z k
The only solutions for i, j, k ≥ 2 are when x, y , z are all powers of
a third word.
Example 4:
XYZ = ZVX : many solutions, but not expressible by formula with
integer parameters.
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Equations in words
More generally, given an equation in words and constants, there is
an algorithm (Makanin’s algorithm) that is guaranteed to find a
solution if one exists.
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Equations in words
More generally, given an equation in words and constants, there is
an algorithm (Makanin’s algorithm) that is guaranteed to find a
solution if one exists.
Satisfiability in PSPACE: Plandowski (2004)
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Equations in words
More generally, given an equation in words and constants, there is
an algorithm (Makanin’s algorithm) that is guaranteed to find a
solution if one exists.
Satisfiability in PSPACE: Plandowski (2004)
Finiteness of solutions: Plandowski (2006) but complete proofs not
yet published.
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Kinds of infinite words
pure morphic: obtained by iterating a morphism
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Kinds of infinite words
pure morphic: obtained by iterating a morphism
Examples:
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Kinds of infinite words
pure morphic: obtained by iterating a morphism
Examples:
the Thue-Morse word, obtained by iterating 0 → 01, 1 → 10
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Kinds of infinite words
pure morphic: obtained by iterating a morphism
Examples:
the Thue-Morse word, obtained by iterating 0 → 01, 1 → 10
the Fibonacci word, obtained by iterating 0 → 01, 1 → 0
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Kinds of infinite words
pure morphic: obtained by iterating a morphism
Examples:
the Thue-Morse word, obtained by iterating 0 → 01, 1 → 10
the Fibonacci word, obtained by iterating 0 → 01, 1 → 0
morphic: obtained by iterating a morphism, then applying a coding
(letter-to-letter morphism)
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Kinds of infinite words
pure morphic: obtained by iterating a morphism
Examples:
the Thue-Morse word, obtained by iterating 0 → 01, 1 → 10
the Fibonacci word, obtained by iterating 0 → 01, 1 → 0
morphic: obtained by iterating a morphism, then applying a coding
(letter-to-letter morphism)
fixed point of uniform morphism: like the Thue-Morse word
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Kinds of infinite words
pure morphic: obtained by iterating a morphism
Examples:
the Thue-Morse word, obtained by iterating 0 → 01, 1 → 10
the Fibonacci word, obtained by iterating 0 → 01, 1 → 0
morphic: obtained by iterating a morphism, then applying a coding
(letter-to-letter morphism)
fixed point of uniform morphism: like the Thue-Morse word
automatic: image, under a coding, of a fixed point of a uniform
morphism
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Other important infinite words
Sturmian words: exactly n + 1 factors of length n
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Other important infinite words
Sturmian words: exactly n + 1 factors of length n
Many other characterizations:
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Other important infinite words
Sturmian words: exactly n + 1 factors of length n
Many other characterizations:
balanced: any two words w , x of length n have δ(x, y ) = 1,
where δ(x, y ) = ||x|1 − |y |1 |
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Other important infinite words
Sturmian words: exactly n + 1 factors of length n
Many other characterizations:
balanced: any two words w , x of length n have δ(x, y ) = 1,
where δ(x, y ) = ||x|1 − |y |1 |
of the form (bα(n + 1) + γc − bαn + γc)n≥1 for real α, γ
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Other important infinite words
Sturmian words: exactly n + 1 factors of length n
Many other characterizations:
balanced: any two words w , x of length n have δ(x, y ) = 1,
where δ(x, y ) = ||x|1 − |y |1 |
of the form (bα(n + 1) + γc − bαn + γc)n≥1 for real α, γ
Episturmian words: natural generalization of Sturmian words to
larger alphabets
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More infinite words
Toeplitz words: generated by starting with a periodic word with
“holes”; then inserting another periodic word with holes into that,
etc.
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More infinite words
Toeplitz words: generated by starting with a periodic word with
“holes”; then inserting another periodic word with holes into that,
etc.
Paperfolding words: generated by iterated folding of a piece of
paper
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More infinite words
Toeplitz words: generated by starting with a periodic word with
“holes”; then inserting another periodic word with holes into that,
etc.
Paperfolding words: generated by iterated folding of a piece of
paper
Xn+1 = Xn aXn
R
168 / 201
More infinite words
Toeplitz words: generated by starting with a periodic word with
“holes”; then inserting another periodic word with holes into that,
etc.
Paperfolding words: generated by iterated folding of a piece of
paper
Xn+1 = Xn aXn
R
Kolakoski’s word: generated by applying a transducer iteratively
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More infinite words
Toeplitz words: generated by starting with a periodic word with
“holes”; then inserting another periodic word with holes into that,
etc.
Paperfolding words: generated by iterated folding of a piece of
paper
Xn+1 = Xn aXn
R
Kolakoski’s word: generated by applying a transducer iteratively
The sequence 1221121221 · · · that encodes its own sequence
of run lengths
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Properties of infinite words
recurrence - every factor that occurs, occurs infinitely often
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Properties of infinite words
recurrence - every factor that occurs, occurs infinitely often
uniform recurrence - recurrent, plus distance between two
consecutive occurrences of the same factor of length n is bounded,
for all n
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Subword complexity
“subword” complexity - given an infinite word w, count the
number of distinct factors of length n in w
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Subword complexity
“subword” complexity - given an infinite word w, count the
number of distinct factors of length n in w
O(n) for automatic sequences
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Subword complexity
“subword” complexity - given an infinite word w, count the
number of distinct factors of length n in w
O(n) for automatic sequences
n + 1 for Sturmian words
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Subword complexity
“subword” complexity - given an infinite word w, count the
number of distinct factors of length n in w
O(n) for automatic sequences
n + 1 for Sturmian words
O(n2 ) for morphic words
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Subword complexity
“subword” complexity - given an infinite word w, count the
number of distinct factors of length n in w
O(n) for automatic sequences
n + 1 for Sturmian words
O(n2 ) for morphic words
A classification of possible growth rates exists
177 / 201
Automatic sequences
A deterministic finite automaton with output (DFAO) is a
6-tuple: (Q, Σ, δ, q0 , ∆, τ ), where ∆ is the finite output
alphabet and τ : Q → ∆ is the output mapping.
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Automatic sequences
A deterministic finite automaton with output (DFAO) is a
6-tuple: (Q, Σ, δ, q0 , ∆, τ ), where ∆ is the finite output
alphabet and τ : Q → ∆ is the output mapping.
Next, we decide on a integer base k ≥ 2 and represent n as a
string of symbols over the alphabet Σ = {0, 1, 2, . . . , k − 1}.
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Automatic sequences
A deterministic finite automaton with output (DFAO) is a
6-tuple: (Q, Σ, δ, q0 , ∆, τ ), where ∆ is the finite output
alphabet and τ : Q → ∆ is the output mapping.
Next, we decide on a integer base k ≥ 2 and represent n as a
string of symbols over the alphabet Σ = {0, 1, 2, . . . , k − 1}.
To compute fn , given an automaton M, express n in base-k,
say,
ar ar −1 · · · a1 a0 ,
and compute
fn = τ (δ(q0 , ar ar −1 · · · a1 a0 )).
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Automatic sequences
A deterministic finite automaton with output (DFAO) is a
6-tuple: (Q, Σ, δ, q0 , ∆, τ ), where ∆ is the finite output
alphabet and τ : Q → ∆ is the output mapping.
Next, we decide on a integer base k ≥ 2 and represent n as a
string of symbols over the alphabet Σ = {0, 1, 2, . . . , k − 1}.
To compute fn , given an automaton M, express n in base-k,
say,
ar ar −1 · · · a1 a0 ,
and compute
fn = τ (δ(q0 , ar ar −1 · · · a1 a0 )).
Any sequence that can be computed in this way is said to be
k-automatic.
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The Thue-Morse automaton
The word t is computed by the following DFAO:
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The Thue-Morse automaton
The word t is computed by the following DFAO:
0
0
1
0
1
1
183 / 201
Robustness
the order in which the base-k digits are fed into the
automaton in does not matter (provided it is fixed for all n);
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Robustness
the order in which the base-k digits are fed into the
automaton in does not matter (provided it is fixed for all n);
other representations also work (such as expansion in
base-(−k));
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Robustness
the order in which the base-k digits are fed into the
automaton in does not matter (provided it is fixed for all n);
other representations also work (such as expansion in
base-(−k));
automatic sequences are closed under many operations, such
as shift, periodic deletion, q-block compression, and q-block
substitution.
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Robustness
the order in which the base-k digits are fed into the
automaton in does not matter (provided it is fixed for all n);
other representations also work (such as expansion in
base-(−k));
automatic sequences are closed under many operations, such
as shift, periodic deletion, q-block compression, and q-block
substitution.
if a symbol in an automatic sequence occurs with well-defined
frequency r , then r is rational.
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Christol’s theorem
Theorem
(Christol [1980]). Let (un )n≥0 be a sequence over
Σ = {0, 1, . . . , p − 1},
where p is
Pa prime. Then the formal power series
U(X ) = n≥0 un X n is algebraic over GF (p)[X ] if and only if
(un )n≥0 is p-automatic.
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Christol’s theorem: example
Let (tn )n≥0 denote the Thue-Morse sequence.
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Christol’s theorem: example
Let (tn )n≥0 denote the Thue-Morse sequence.
Then tn = sum of the bits in the binary expansion of n, mod 2.
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Christol’s theorem: example
Let (tn )n≥0 denote the Thue-Morse sequence.
Then tn = sum of the bits in the binary expansion ofPn, mod 2.
Also t2n ≡ tn and t2n+1 ≡ tn + 1. If we set A(X ) = n≥0 tn X n ,
then
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Christol’s theorem: example
Let (tn )n≥0 denote the Thue-Morse sequence.
Then tn = sum of the bits in the binary expansion ofPn, mod 2.
Also t2n ≡ tn and t2n+1 ≡ tn + 1. If we set A(X ) = n≥0 tn X n ,
then
A(X ) =
X
n≥0
t2n X 2n +
X
t2n+1 X 2n+1
n≥0
192 / 201
Christol’s theorem: example
Let (tn )n≥0 denote the Thue-Morse sequence.
Then tn = sum of the bits in the binary expansion ofPn, mod 2.
Also t2n ≡ tn and t2n+1 ≡ tn + 1. If we set A(X ) = n≥0 tn X n ,
then
A(X ) =
X
t2n X 2n +
=
X
tn X 2n + X
n≥0
n≥0
X
t2n+1 X 2n+1
n≥0
X
n≥0
tn X 2n + X
X
X 2n
n≥0
193 / 201
Christol’s theorem: example
Let (tn )n≥0 denote the Thue-Morse sequence.
Then tn = sum of the bits in the binary expansion ofPn, mod 2.
Also t2n ≡ tn and t2n+1 ≡ tn + 1. If we set A(X ) = n≥0 tn X n ,
then
A(X ) =
X
t2n X 2n +
=
X
tn X 2n + X
n≥0
n≥0
X
t2n+1 X 2n+1
n≥0
X
n≥0
tn X 2n + X
X
X 2n
n≥0
= A(X 2 ) + XA(X 2 ) + X /(1 − X 2 )
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Christol’s theorem: example
Let (tn )n≥0 denote the Thue-Morse sequence.
Then tn = sum of the bits in the binary expansion ofPn, mod 2.
Also t2n ≡ tn and t2n+1 ≡ tn + 1. If we set A(X ) = n≥0 tn X n ,
then
A(X ) =
X
t2n X 2n +
=
X
tn X 2n + X
n≥0
n≥0
X
t2n+1 X 2n+1
n≥0
X
tn X 2n + X
n≥0
X
X 2n
n≥0
= A(X 2 ) + XA(X 2 ) + X /(1 − X 2 )
= A(X )2 (1 + X ) + X /(1 + X )2 .
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Christol’s theorem: example
Let (tn )n≥0 denote the Thue-Morse sequence.
Then tn = sum of the bits in the binary expansion ofPn, mod 2.
Also t2n ≡ tn and t2n+1 ≡ tn + 1. If we set A(X ) = n≥0 tn X n ,
then
A(X ) =
X
t2n X 2n +
=
X
tn X 2n + X
n≥0
n≥0
X
t2n+1 X 2n+1
n≥0
X
tn X 2n + X
n≥0
X
X 2n
n≥0
= A(X 2 ) + XA(X 2 ) + X /(1 − X 2 )
= A(X )2 (1 + X ) + X /(1 + X )2 .
Hence (1 + X )3 A2 + (1 + X )2 A + X = 0.
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Open Problems
Is the set of primitive words over {0, 1} context-free? (Almost
certainly not.)
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Open Problems
Is the set of primitive words over {0, 1} context-free? (Almost
certainly not.)
What are the frequencies of letters in Kolakoski’s word? Do
they exist? Are they equal to 1/2?
198 / 201
Open Problems
Is the set of primitive words over {0, 1} context-free? (Almost
certainly not.)
What are the frequencies of letters in Kolakoski’s word? Do
they exist? Are they equal to 1/2?
Nivat’s conjecture: extension of periodicity to 2-dimensional
arrays
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Open Problems
Is the set of primitive words over {0, 1} context-free? (Almost
certainly not.)
What are the frequencies of letters in Kolakoski’s word? Do
they exist? Are they equal to 1/2?
Nivat’s conjecture: extension of periodicity to 2-dimensional
arrays
Is there a wordPover aP
finite subset of N that avoids xx 0 with
0
|x| = |x | and
x=
x 0?
200 / 201
For Further Reading
M. Lothaire, Combinatorics on Words, Cambridge, 1997
(reprint)
M. Lothaire, Algebraic Combinatorics on Words, Cambridge,
2002
M. Lothaire, Applied Combinatorics on Words, Cambridge,
2005
V. Berthé, M. Rigo, Combinatorics, Automata, and Number
Theory, Cambridge, 2010
J.-P. Allouche and J. Shallit, Automatic Sequences: Theory,
Applications, Generalizations, Cambridge, 2003.
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